Microwave Transmission Lines

Abstract
This document presents an introduction to the basics of microwave transmission lines. It is important to
understand the principles underlying the propagation and transmission of high-frequency signals, which
are vital in areas such as communications circuit design as well as in high-frequency processor cores, and it
is a well known fact that contemporary processors are clocked by frequencies as high as 3GHz! Designs at
such high frequencies require careful consideration so as to minimize losses and to ensure maximum power
transmission. This document starts by giving an insight into the basics of transmission lines and wave
propagation theory. This is then followed by different transmission line technologies adopted in modern
electronic systems for fabrication.

Usually. it makes use of communication frequencies of about 1800MHz. why not use much higher frequencies? Above microwaves. If antenna sizes can be reduced considerably by using
high frequenices. waves with wavelengths ranging from
as low as a few millimeters to almost a metre are classified as microwaves. and how are they advantageous? The answer is that
most of modern electronic communication engineering make use of microwaves. Would these make good candidates for
wireless communication. For proper wireless transmission
and reception. Moreover. there are the infra-red and
visible light spectra. especially mobile phones.
and the antenna size is usually determined by the wavelength λ. these waves would be afftected by the earth’s
atmosphere. what do microwaves have that makes them suitable for use in comminication engineering? Let us consider for example. A good antenna size approximation is
λ/4. every device requires a transmission/reception antenna. X-rays and gamma rays. where users are often inside their houses. say for example 100Hz?
This leads to another interesting question. are detrimental for health as they are ionizing radiations. Then again.e. This also means that for satellite communications. and for an 1800MHz wave.2cm. an indispensable communication tool for all. What do these have to offer. Supposing that a mobile uses the GSM1800
band. even non-ionizing waves such as infrared rays are easily affected by atmospheric
constituents and do not have obstruction penetration strength. tuned to the frequency of operation. surrounded
by thick walls?
1
.
a mobile phone.Chapter 1
Introduction
Microwaves are a part of the electromagnetic spectrum. A very important question is the reason behind
studying microwaves. there are ultraviolet rays. Conventional definition for the
microwave frequency range is from 300MHz − 300GHz. i. and even above. Many of the
high-frequency waves such as the UV rays and above. the antenna size required would be around 4. What would be the
required antenna size if the transmission frequency would be low.

or in general. which is a function of both time t and the position on the line x.
i.2. Hence.
In such cases.
Vs (t = t0 + tp .. x = 0) = V2 . x = 0) = V1 . wherein the
laws of lumped circuit theory can be applied. i. Voltages V1 and V2 are likely to e largely different.1. Vs (t = t0 . after the line propagation delay time. it is seen that the voltage varies along the length of the line when
the frequency is high. the lumped circuit theorems at low frequencies cannot directly be applied
to analyzing circuits at high frequencies.e. Assume x = 0 denotes the source end and x = l the load end. Consider the source sine wave traveling
along the length l with a wave velocity v. is true for low frequencies only. Define the propagation time from source to load tp as
tp =
v
l
Define a voltage along the line. the voltage across the load Vo (t) will equal the source voltage i.
l
V s (t)
R L V o (t)
Figure 1.1
The High Frequency Circuit Analysis Problem
In this section. the time period T is small. Vo (t) = Vs (t).
2
. The voltage at any given point on the line is now a function of
time as well as space (or position). Consider the source voltage Vs (t)
waveform shown in Figure 1. Consider the source voltage at t = 0 i.
Consider a circuit with a source Vs (t) and load RL .
The starting method of analyzing high frequency transmission lines is to consider a small section of the
line. An initial look at this circuit shows that
at any instant of time t.e. when the wave dimensions become comparable to the dimensions of
the circuit components.e. Clearly.1: Simple source-load circuit
This approximation however. Then. where
0 ≤ x ≤ l. if the wave frequency is high. where the voltage is assumed not to change significantly over the length of the section.
the load voltage will equal V1 only when t = t0 + tp .e. But. we will see the method of analysis for low-frequency circuits and from thereon. the wire length l has absolutely no role is determining Vo (t) (assuming l is small enough).1. examine
why this cannot be used at high frequencies. The connection between them is by
means of two conductors (whose resistance is assumed to be low). then. as in Figure 1.

while at a distance of ∆x
towards the right are V (x + ∆x) and I(x + ∆x). the conductors
will have some resistance.2
Voltage and Current along a Transmission Line
Let us now study the transmission line behaviour to an input. applying
Kirchhoff’s Voltage Law (KVL) in the loop. G. Consider an
infinitesimally small section of the transmission line ∆x. in greater detail. the actual resistance. The
voltage and current at the left end of the line are denoted as V (x) and I(x). Irrespective of the kind of signal traveling through the line.
capacitance etc. Let R denote the resistance of the line per unit length. The section of such a size can then be expressed as an R − L − G − C lumped circuit as shown
in Figure 1. we have
V (x + ∆x) = V (x) − I(x)(R + jΩL)∆x
3
. Due to the traveling sine wave.
Assume that the frequency of operation is f . ∆x. The magnetic
field leads to a distributed inductance along the line. for this section of the transmission line will be the product of these parameters and the
length in consideration.
Since we have assumed R. while a mutual electric field will interact between the two conductors. L. which may have a parasitic conductance component. Now. Also. i. a time-varying magnetic field will be generated around
each conductor. Denote the per unit length inductance and capacitance as L and C respectively. where the voltage variation over the length ∆x is
negligible. Consider a sinusoidal
signal traveling along the line. C to be the per unit length line parameters. and the angular frequency Ω = 2πf .e. which is assumed to be infinitesimally small.1. this two-conductor
connection is separated by a dielectric. denoted per
unit length as G.3.V s (t)
V1
V2
tp
to
t
T
Figure 1.2: Sinusoidal source voltage
1. while the interacting electric field leads to a mutual
capacitance.. Note that we deal with
the positional voltages and currents in this circuit as a consequence of the discussion in Section 1.

differentiate both (1.3: Small section of a transmission line. The two differential
equations now become
d2 V
= γ2V
(1. we get
dI
= −(G + jΩC)V
dx
(1.5). Let us see why.4)
dx2
Let γ 2 = (R+jΩL)(G+jΩC).2)
We now have two differeintial equations in two variables. V and I.1)
Similarly.7)
.
Consider the voltage equation (1. which can be solved independently. We now have two equations
d2 V
= (R + jΩL)(G + jΩC)V
dx2
(1.R∆x
I(x)
V(x)
L∆x
I(x+∆x)
G∆x
C∆x V(x+∆x)
Figure 1. These equations are now
simplified. The significance of the term γ will be dicussed later.5) and (1. To make them easier to solve.lumped circuit equivalent
V (x + ∆x) − V (x)
= −I(x)(R + jΩL)
∆x
taking the limiting case of ∆x → 0
∴
V (x + ∆x) − V (x)
dV
=
= −(R + jΩL)I
∆x→0
∆x
dx
lim
(1.6)
2
dx
Equations (1.5)
dx2
d2 I
= γ2I
(1.2) by x again.6) are called the wave equations.1) and (1.3)
d2 I
= (R + jΩL)(G + jΩC)I
(1. by applying Kirchhoff’s Current Law (KCL) at the right node. which are ordinary second order differential equations. Its solution is of the form
V (x) = V1 e−γx + V2 eγx
4
(1.

t). which is the real
part of the second term of the solution for V (x. for simplicity.9). Thus. Consider. is a complex quantity. comprising of a sine and a cosine component. the results are as shown in Figure 1. let
γ = α + jβ. V1 cos(Ωt − βx). t) = V1 e−γx ejΩt + V2 eγx ejΩt
(1. This indicates that when a transmission line is excited with
an AC input.from source to load and vice-versa. indicating
the wave “propagation” in the positive x direction. the current will
also propagate in two directions!
5
. Thus γ in general. there are two waves traveling. for different values of x). As the value of x increases.9)
This gives an interesting result. the term V2 cos(Ωt + βx). i. This quantity is complex. the point A shifts to the right. the complete
solution for V (x) is a function of both x and time t.
which can be represented by a complex exponential ejΩt . However. Consider a sinusoidal input to the transmission line. the complete voltage solution on the line
is
V (x.e. Similarly. If this term is now evaluated for various values of x in increasing
order.
A
x = xo
A
x = x1 > xo
A
x = x2 > x1
Figure 1.4. Consider the real
part of this term. α = 0.4: Wave propagation as a function of x
Observe the point A on the wave. Let us focus on the first term of (1. Therefore.The solution for V (x) is a function of the position on the transmission line alone. Likewise. t) = V1 e−αx ej(Ωt−βx) + V2 eαx ej(Ωt+βx)
(1. indicates a wave traveling in the negative x direction
(work it out yourself.e. t) can now be represented as
V (x.8)
Recall that γ 2 = (R + jΩL)(G + jΩC). V (x.
i. V1 ej(Ωt−βx) .

and applications
of transmission lines. we again select
the voltage wave component that is traveling toward the positive x direction as we had in Section 1. the wavelength of the propagating wave on the
6
. γ is a quantity that determines the wave propagation. t) = V1 e−αx ej(Ωt−βx) + V2 eαx ej(Ωt+βx)
I(x. it is
called the attenuation constant. the sine wave amplitude (envelope) changes exponentially over the distance x. Let us investigate the quantity γ in more detail. the sine wave peak would be V1 .e. and that waves propagate in two directions. such as the line
behaviour with different kinds of load. Hence. hence β is
called the phase constant. the complete solution for the voltage and current along a transmission
line is given as
V (x.Chapter 2
Transmission Line Characteristics
In the previous chapter.1
Wave Propagation in Transmission Lines
We have seen that in general. i. Note that a distance of λ.
2. we will study the transmission line characteristics in greater detail. as it exists due to the mere presence of the input
ejΩt itself. we had an initial look at the lumped-component sectional representation of
a transmission line. Let us now look into the quantities α and β more closely. As α causes change in wave amplitude over the length of the transmission line.e. once the line is excited with a voltage
input. This is
then. Due
to a nonzero α. The term βx in (2. issues of losses and different techniques for analysis.1)
Note that γ = α + jβ.
V1 e−αx cos(Ωt − βx). it is called the
propagation constant.1) denotes a phase component of the wave.2. t) = I1 e−αx ej(Ωt−βx) + I2 eαx ej(Ωt+βx)
(2. Now what does α indicate? Had α been zero. i. Suppose γ = 0. To do this. Thus clearly. a trivial case of a non-propagating sinusoidal wave. In this chapter. depending
on the value of α.

which
is the distance along the transmission line. However. it is treated in the same way as voltage or current gain and is sometimes expressed in
decibels (dB/m). equation (1. Considering moving towards the load a positive x direction. αx = 1. since α in some sense. Likewise.68dB. at which the wave amplitude becomes 1/e times the starting
amplitude (at x = 0). Thus. microwave engineers deal with a term named “effective travel distance”.
2.2
Line and Load Impedance
Let us consider the positional solutions for the voltage and current on the transmission line. at this point. named Neper (N p). i. We
define here a new unit.7) can be re-written
for both voltage and current as
V (x) = V+ e−γx + V− eγx
I(x) = I+ e−γx + I− eγx
7
(2. α would have the units m−1 . Clearly. The effective dB gain is 20 log10 e−1 = 8. for α > 0.68dB. for x = λ.1: Propagating wave envelope for α > 0
As seen in Figure 2. α is often expressed in the unit
N p/m−1 . βλ = 2π.2)
What do you think will be the units for α and β? A first look tells us that since the term αx is
dimensionless (why?).e.3)
.1. This indicates the absolute attenuation affecting the
wave per unit length of the transmission line.transmission line would mean a phase change of 2π. Thus. Hence
β=
2π
λ
(2. V (x)
and I(x). Now we know that the voltage and current have two components propagating in opposite
directions. where 1N p = 8. the wave amplitude decreases exponentially as it propagates towards
the load. This clarifies the reaon why α is called the attenuation constant. the units for β would be rad/m. Often.
wave envelope
ZL
direction of wave propagation
Figure 2. relates to a change in
voltage or current.

Thus. which is
dV
= −(R + jΩL)I
dx
s
p
R + jΩL
I
= − (R + jΩL)(G + jΩC)
G + jΩC
q
R+jΩL
has the units of impedance.e. i.3) gives
us
V (l) = V+ eγl + V− e−γl
I(l) = I+ eγl + I− e−γl
8
(2. substituting x = −l in (2.As seen in (2. the current equation may
also be expressed in terms of V+ and V− . Let us see how. Consider equation (1. the impedance Z0 seems to govern the line voltage and current much like Ohm’s Law. We now have
dV
= −γZ0 I
(2.5)
In some sense. Z0 is thus called the characteristic impedance of the transmission line and is a very important quantity governing microwave-based
designs. i. but rather. L. It will be soon clear that the load connected at the end of the line plays
a very important factor in determining the voltage and current variation along the length of the line. We will make use of the knowledge of γ as a
function of line parameters. Let
us therefore. V− . there are four unknowns.3) as per the results obtained in (2.e.3).4)
dx
Now let us differentiate the voltage expression in (2. I+ and I− . a
distributed impedance that is characteritic to the line parameters themselves. The term G+jΩC
Z0 . so let us call it
Observe the above expression carefully.
Let us now change our analysis slightly. and also on the frequency
of operation Ω. consider a length axis l.1). However. V+ . we have now focused more on the transmission line
parameters than anything else.6)
. So far. G and C.4). γ 2 = (R + jΩL)(G + jΩC). Thus.
−γZ0 I(x) = −γV+ e−γx + γV− eγx
∴ I(x) =
1
(V+ e−γx − V− eγx )
Z0
(2. Z0 is an impedance not physically present anywhere on the line.
Z0 is seen to depend on the per unit length line parameters R. where l = 0 is the position of the load and movement twoards the
voltage source indicates movement along the positive l direction.

the reflection coefficient at any point along the line is the ratio of the backward moving wave to
the forward moving wave in terms of voltage magnitude.10) by
ΓL =
ZL − Z0
ZL + Z0
9
(2.5). we have
V (l) = V+ eγl + V− e−γl
1
(V+ eγl − V− e−γl )
I(l) =
Z0
(2. In
some sense. as follows
Γ(l) =
V− e−γl
V− −2γl
=
e
V+ eγl
V+
(2. where
V+ eγl + V− e−γl
Z(l) = Z0
V+ eγl − V− e−γl
At l = 0.Using the results obtained in (2. the voltage
component V− e−γl .10)
where ΓL is the reflection coefficient at the load end.towards
the load and source respectively. Hence. Now.
ZL = Z0
1 + ΓL
1 − ΓL
(2.9)
Thus. Thus. without any power reflected
back from the load. reflection towards the source can degrade the performance of the source
itself over time. called
voltage reflection coefficient. taking
V+ common from both numerator and denominator. Intuitively.8)
Recall that the quantities V+ eγl and V− e−γl denote wave propagation in opposite directions. this quantity is a voltage “reflected” back from the load.7)
The impedance at any point on the line is thus given by
Z(l) =
V (l)
V+ eγl + V− e−γl
= Z0
I(l)
V+ eγl − V− e−γl
(2. at that point. ΓL can be expressed from (2. we aim to have the reflection coefficient Γ as small as possible. Consider equation
(2. Moreover. i.8).11)
. we can say that the voltage component V+ eγl travels towards
the load due to excitation from the source at the other end of the transmission line. which travels back to the source. we have the load impedance Z(0) = ZL = V (0)/I(0). at the load. Let us define a quantity Γ. However. is not what one would have expected to exist. The original idea of designing a
transmission line is to transmit the entire power from the source to the load. at l = 0.e.

which is true since the maximum power that can be reflected back
to the source can equal the input power itself (principle of conservation of energy).
γ=
p
(R + jΩL)(G + jΩC)
To have α = 0. clearly the load impedance has to be equal to the line characteristic impedance.1
Transmission Line Losses
We have seen that the quantity γ governs the wave propagation characteristics in terms of both amplitude and phase.
2. We would like the transmission line to be a lossless
line. Here. ZL = Z0 . we have seen in section 2.Equation (2. as a result. decreases) exponentially as we move towards the load. Thus. we look briefly into
certain related issues that are important for engineers who wish to design transmission lines at very high
frequencies. we
now have. for the line to be lossless. To ensure zero reflected
power. i. It is clear that ΓL (subsequently. It
is evident that the quantity α governs the power delivered to the load.e. As γ = α + jβ.3.1 that α affects the wave amplitude variation in
the line and β is related to the phase of the sinusoid.the wave propagation.3
Design Issues in Transmission Lines
So far. it must have α = 0. What happens when the load is (a) an open circuit and (b) a
short circuit?
2. i.e. we have studies some important characteristics in a transmission line.11) gives some interesting information.
√
γ = jΩ LC
√
=⇒ β = Ω LC
10
(2. line
and load impedances and how they affect the wave characteristics in the line. If α 6= 0.12)
. we need γ = jβ. This
directly indicates a loss. we have seen that the
wave amplitude changes (more specifically. This can only happen when R = G = 0.
What is the physical implication of all this? How does the fact that a lossless line having α = 0
translate physically? Let us revert to the expression for the propagation constant γ. the power that reaches the load should be the same as the power generated by the source. Such
a condition is known as a matched condition and is critical in ensuring that the load absorbs all the
indcident power and reflects back nothing. the Γ at any point
l) takes values such that 0 ≤ |Γ(l)| ≤ 1. the voltage amplitude that reaches the load is not the same as that generated by
the source. We know that. With this.

12).
Γ(l) = ΓL e−2γl
Since the term e−2γl is difficult to be made zero (as neither l nor γ is zero). which is given by (2.14)
This indicates the physical properties of a lossless line.
By definition.9). The matched condition requires that Γ = 0. in order
to minimize reflections from the load. there are voltage and current waves
traveling on the line. From (2. Thus. β = 2π/λ. ΓL is the reflection coefficient at the load. where the concept of reflection coefficient Γ was introduced. Thus.We have seen that.
these are lossy components. The waves reflected back to the source are highly undersirable as they tend to
load the source and reduces its life. as
ΓL =
ZL − Z0
ZL + Z0
11
. Since the AC impedance of R and G are frequency-independent. Resistance and conductance are lossy elements
and must be zero to minimize the losses. we have.2.3.
2. which can be expressed as
v=√
1
LC
(2. The waves traveling back to the source are
termed as reflected waves. we need to make ΓL = 0.2
Matching in Transmission Lines
We have discussed that when a line is excited by an AC source. we get
√
2π
= 2πf LC
λ
The wave velocity v is the product of frequency and wavelength. It is
the ratio of the amplitudes of the reverse-going wave to the forward-going gave.
Recall our discussion in section 2. since both R = G = 0.
Z0 =
r
L
C
(2. at any point l along the
line. the second important design consideration is matching. in both directions.to and from the load.11). unlike the line inductance and capacitance.13)
For a lossless line the characteristic impedance Z0 is purely real. Substituting in (2.

then we can calculate the impedance Z at any other point on the line l.if we know the characteristic impedance Z0 and the impedance
at any point on the line l1 . Thus. Substituting.3
Line Impedance Revisited
We have already derived the expressions for voltage and current as a function of position l on the
transmission line.15)
This is a very important result and is known as the matched condition.
12
. we get
Z(l) = Z0
eγl + ΓL e−γl
eγl − ΓL e−γl
Also.
this relation is known as the impedance transformation relation and is a very handy tool.16)
Equation (2. It indicates that if we know the characteristic impedance
Z0 and the load impedance ZL .16) is a very important result. there will
be no reflected wave from the load to the source. Putting this we get
Z(l) =
=
=
ZL −Z0 −γl
e
ZL +Z0
Z0 γl ZL −Z0 −γl
e − ZL +Z0 e
(ZL + Z0 )eγl + (ZL − Z0 )e−γl
Z0
(ZL + Z0 )eγl − (ZL − Z0 )e−γl
ZL (eγl + e−γl ) + Z0 (eγl − e−γl )
Z0
ZL (eγl − e−γl ) + Z0 (eγl + e−γl )
eγl +
Expressing the sum of exponentials as hyperbolic cosine and sine functions. ΓL = (ZL − Z0 )/(ZL + Z0 ).
2. this equation becomes
Z(l) = Z0
ZL cosh γl + Z0 sinh γl
ZL sinh γl + Z0 cosh γl
(2. this can be generalized even further.8) as
Z(l) = Z0
V+ eγl + V− e−γl
V+ eγl − V− e−γl
We know that ΓL = V+ /V− . We would like to design lossless (zero attenuation) and matched transmission lines.3. then we can find the impedance at any other point on the line l2 6= l1 ! Hence. the impedance at any point l on the line is given by (2. Under this condition. In
fact. The matched condition is somewhat analogous to the
maximum power transfer theorem as in circuit theory and is a highly desirable condition in the design of
transmission lines.ΓL = 0 would imply that
ZL = Z0
(2.

let us now examine some interesting properties of a lossless transmission line.Let us now bring in a new term called normalized impedance. which is. The normalized impedance is a useful term. Consider the impedance Z(l) whose normalized impedance is
expressed as
Z(l)
z¯(l) =
(2. we have
j z¯L sin βl + cos βl
z¯L cos βl + j sin βl
1
=⇒ z¯(1 + λ/4) =
z¯(l)
z¯(l + λ/4) =
We can thus state the first result as
Result 1 The normalized line impedance inverts itself every λ/4 distance along the line. Putting this in the above expression. wherein the impedance at any point
on the line is taken with reference to Z0 . In terms of this.
Doing so gives us the impedance transformation relation for lossless lines. whose significance shall
be discussed later.20)
. say. we have
z¯L cos(βl + π/2) + j sin(βl + π/2)
j z¯L sin(βl + π/2) + cos(βl + π/2)
−¯
zL sin βl + j cos βl
=
j z¯L cos βl − sin βl
z¯(l + λ/4) =
Multiplying both numerator and denominator by −j. z¯(l). the z¯L term is unitless. the impedance transformation relation can be written as
z¯(l) =
z¯L cosh γl + sinh γl
z¯L sinh γl + cosh γl
(2. as
z¯(l) =
z¯L cos βl + j sin βl
j z¯L sin βl + cos βl
(2. The normalized impedance at
this point is then
z¯L cos β(l + λ/4) + j sin β(l + λ/4)
z¯(l + λ/4) =
j z¯L sin β(l + λ/4) + cos β(l + λ/4)
We know that β = 2π/λ. where λ is the operating wavelength.18). Let us now move along the
length of the line by a distance λ/4.17)
Z0
As is clear. we can then replace γ = α+jβ. In (2. Suppose
we know the normalized impedance on a line at point l.19)
Based on this.
13
(2.18)
Let us now focus our discussion on lossless transmission lines.

yet indispensable result.
All line impedances are with respect to the characteristic impedance Z0 . Hence. in many discussions in
future. let us move a distance of λ/2 along the transmission line. these become
V (l) = V+ ejβl + V− e−jβl
1
I(l) =
(V+ ejβl − V− e−jβl )
Z0
(2.22)
. These are given by (2.
As far as matching is concerned.21)
This set of lines may be rewritten by taking the V+ ejβl term common.
V (l) = V+ eγl + V− e−γl
1
I(l) =
(V+ eγl − V− e−γl )
Z0
For lossless lines. The normalized impedance is then
z¯L cos(βl + π) + j sin(βl + π)
j z¯L sin(βl + π) + cos(βl + π)
z¯L cos βl + j sin βl
=
j z¯L sin βl + cos βl
z¯(l + λ/2) =
= z¯(l)
This brings us to the second result.7).Now. the absolute load impedance does not have much meaning by itself.4
Wave Patterns in Transmission Lines
Let us now revisit the solutions for voltage and current on transmission lines as a function of the position
l.
2. This is done in order to find the
maximan and minima of the voltage and current along the line.
Result 2 The line impedance repeats itself every λ/2 distance along the line.
V (l) = V+ ejβl (1 + ΓL e−j2βl )
V+ jβl
I(l) =
e (1 − ΓL e−j2βl )
Z0
14
(2. we will find ourselves discussing more with the normalized impedance rather than the absolute
impedance.
Result 3 The normalized load impedance for a matched load is unity. This brings us to another obvious. Since we already know the definition of
reflection coefficient Γ.

Such comptutations can be done by means of the impedance transformation relation. α = 0. that 0 ≤ |ΓL | ≤ 1.
limit our discussion to lossless transmission lines i. consecutive maxima or minima are apart from each other by a distance of λ/2. For instance. the Zmax is simply equal to the VSWR! Thus.
the load impedance can be caluclated provided we know the impedance at some other point on the line. We shall. is rather tedious and non-intuitive. In this
section. we can
very easily compute the maximum and minimum impedance along the transmission line. which will aid us in solving
problems on transmission lines. i.
Also. this analysis
shows us the pattern of the standing waves along the line.27)
z¯ + 1
Recall that Γ is a complex quantity.27) indicates that every value of z¯ can be
16
. This tool was developed by Philip H. denoted as z¯. At the point of
maximum impedance.5
The Smith Chart
We have seen that the impedance at any point on the line can be calculated provided we know the load
impedance and the characteristic impedance. For a perfectly matched line. Can you prove this?
2.23) and (2. Smith in 1939 and is still widely in use. though straightforward.
Consider the normalized impedance at any point on the line.e.26)
And in terms of normalized impedance. This quantity VSWR is merely an extension to our knowledge of the reflection
coefficient. We can also do these operations the other way round. and the expression in (2. as
ρ=
1 + |ΓL |
|Vmax |
=
|Vmin |
1 − |ΓL |
(2. From
(2. Voltage and current waves are out of phase
by 180◦ (from the fact that voltage maxima/minima coincide with current minima/maxima respectively). the voltage magnitude peaks and the current magnitude reaches a minimum.24).Define a Voltage Standing Wave Ratio (VSWR) ρ. which means that 1 ≤ ρ ≤ ∞.e. the
load VSWR should be unity. we shall study a simple graphical or figurative tool. but is a highly useful tool for measurements at microwave frequencies.25)
We know by now. the Smith Chart. The reflection coefficient
at this point is then given by
z¯ − 1
Γ=
(2. for now. the computation of impedances
using the transformation relation. you can show that
Zmax = ρZ0
Z0
Zmin =
ρ
(2. However.

Consider the point
Γ = 0. i. As an exrecise.they are calibrated much like a stationery
graph sheet. would be the point where the
r = 0. x = 0 circles intersect. as shown
in Figure 2. z¯ = ∞ would be the point where the
r = ∞. You can verify this easily by considering the open. the corresponding impedance is equal
to the characteristic impedance. Thus. A short-circuit impedance.5: Smith Chart as a superposition of resistance and reactance circles
Let us now see some interesting results from the Smith Chart in Figure 2. Figure 2. point out the resistance and reactance
circles for discrete values. and an open-circuit impedance.
Commonly used Smith Charts don’t just have a few circles. the Smith Chart is obtained
as a result.with accurately labeled resistance and reactance circles.e. and one full circumference implies a
λ/2 movement. Simply by superimposing the resistance and reactance circles. i.
20
. Combining
the results of the constant resistance and reactance circles.3 that the normalized line impedance inverts itself after every
λ/4 length along the line and repeats every λ/2 length along the line.3. Figure 2.6 shows the results. x = ∞ circles intersect.4 that a
clockwise movement on the Γ-plane corresponds to movement towards the source and vice-versa (as per
our conventions on length l). z¯ = 0. we see that every impedance is uniqeuly identitfied by intersection
of a constant resistance and a constant reactance circle. The outermost circle points to r = 0.
Figure 2.e.and short-circuit points on the line. This means that half the Smith
Chart circumference corresponds to a movement of λ/4 on the line. i.
Recall from our discussion in section 2.e.5.5. Recall from our discussion in section 2. This would mean that this point corresponds to z¯ = 1.7 shows a commercially
available Smith Chart that is commonly used to solve transmission line problems.those on the negative half of the Γ-plane correspond to the capacitive reactance component. we get what is called the Smith Chart.